Question: A few families took a trip to an amusement park together. Tickets cost $$5.50$ each for adults and $$3.00$ each for kids, and the group paid $$52.00$ in total. There were $6$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Solution: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${5.5x+3y = 52}$ ${x = y-6}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-6}$ for $x$ in the first equation. ${5.5}{(y-6)}{+ 3y = 52}$ Simplify and solve for $y$ $ 5.5y-33 + 3y = 52 $ $ 8.5y-33 = 52 $ $ 8.5y = 85 $ $ y = \dfrac{85}{8.5} $ ${y = 10}$ Now that you know ${y = 10}$ , plug it back into ${x = y-6}$ to find $x$ ${x = }{(10)}{ - 6}$ ${x = 4}$ You can also plug ${y = 10}$ into ${5.5x+3y = 52}$ and get the same answer for $x$ ${5.5x + 3}{(10)}{= 52}$ ${x = 4}$ There were $4$ adults and $10$ kids.